Question

lf the plane $lx+my+nz=p$ touches the sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ and $(lu+mv+nw+p{)}^2=k(u^2+v^2+w^2-d)$, then $k=$

• A. $l+m+n$
• B. $\sqrt{l^2+m^2+n^2}$
• C. $l^2+m^2+n^2$
• D. $\sqrt{l+m+n}$

Answer 1

The correct option is C $l^2+m^2+n^2$

If line $lx+my+nz=p$ touches the sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$, then distance of line from centre of the sphere is same as its radius.
$\Rightarrow \left|\frac{-lu-mv-nw-p}{\sqrt{l^2+m^2+n^2}}\right|=\sqrt{u^2+v^2+w^2-d}$
$\Rightarrow (lu+mv+nw+p{)}^2=(l^2+m^2+n^2\left)\right(u^2+v^2+w^2-d)$
Comparing it with $(lu+mv+nw+p{)}^2=k(u^2+v^2+w^2-d)$, we get $k=l^2+m^2+n^2$